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Find all positive integers x,y and all prime numbers p for which holds that $p^x-y^p=1$.

For now I only tried Fermat's little theorem.

For now I have the solutions $(x,y,p)={(0,0,k),(1,1,2),(2,2,3)}$, where k is any prime number.

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  • $\begingroup$ I think this works $p=3$ and $x=y=2$. and I claim that is the only solution. $\endgroup$
    – marya
    Nov 18, 2014 at 19:55
  • $\begingroup$ What about x=y=1 and p=2 and x=y=0 and any prime p ? $\endgroup$
    – CryoDrakon
    Nov 18, 2014 at 19:57
  • $\begingroup$ I think this works $p=3$ and $x=y=2$. and I claim that is the only solution. $\endgroup$
    – marya
    Nov 18, 2014 at 19:57
  • $\begingroup$ Note that $p^x - 1 \geq 0 \Longrightarrow |y| = \sqrt[p]{p^x-1}$. $\endgroup$
    – Tacet
    Nov 18, 2014 at 20:01
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    $\begingroup$ possible duplicate of Solving $x^p + y^p = p^z$ in positive integers $x,y,z$ and a prime $p$ $\endgroup$ Nov 19, 2014 at 9:31

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So I have been thinking and and I think I found a way to solve it.

If $x=0$ then it will implicates that y has to be 0 too for every prime p. That goes also the other way around, so we can say that $x,y>0$

Let's first see for p=2 what we have: $2^x-y^2=1$

It's obvious that y has to be an odd number so we can write $y=2n+1$, where n is an non-negative integer. Putting that into the equation gives us $2^x-(4n^2+4n+1)=1$. That is equivalent to $2^{x-1}-2n(n+1)=1$. For $n>0,x>1$ we have that the left side is even and the right is odd which is impossible so it has to be that $n=0$ and $x=1$. For those numbers we have that $y=1$.

Now let's see for $p\ge2$. Now we also have that $p$ is always an odd number.

http://www.artofproblemsolving.com/Resources/Papers/LTE.pdf So this is an article about the LTE Lemma which I am going to use here. $x=v_p(p^x)=v_p(y^p+1)=v_p(y+1)+v_p(p)=v_p(y+1)+1$

$v_p(y+1)=x-1 \Rightarrow p^{x-1}\|y+1$

So we can say there exists an $\alpha$ so $y+1=p^{x-1}\alpha, p\nmid\alpha$.

$p^x=y^p+1=(y+1)(y^{p-1}-y^{p-2}+y^{p-3}-y^{p-4}+...+y^4-y^3+y^2-y+1)$

Let $(y^{p-1}-y^{p-2}+y^{p-3}-y^{p-4}+...+y^4-y^3+y^2-y+1)=t$.

$p^x=p^{x-1}\alpha t\Rightarrow p=\alpha t$, and bcs $p\nmid\alpha$ we have that $t=p$ and $\alpha=1$.

Now we have:

$p(1-p^{x-2})=p-p^{x-1}=(y^{p-1}-y^{p-2}+y^{p-3}-y^{p-4}+...+y^4-y^3+y^2-y+1)-(y+1)=(y^{p-2}(y-1)+y^{p-4}(y-1)+...+y^3(y-1)+y^2-2y)=((y-1)(y^{p-2}+y^{p-4}+...+y^5+y^3+y(y-2))$

$p(1-p^{x-2})=((y-1)(y^{p-2}+y^{p-4}+...+y^5+y^3+y(y-2))$

If $y=1$ Then we have that $p^{x-1}=y+1=1+1=2\Rightarrow p=2,x=2$. Bcs $p\ge3$ those solutions are not correct.

If $y=2$ we have that $p^{x-1}=y+1=2+1=3 \Rightarrow p=3,x=2$, which is one of the solutions.

If $y>2$ we have that the right side is positive, so the left has to be psitive too: $p-p^{x-1}>0\Rightarrow p^1>p^{x-1}$. From here we see that it has to be that $1>x-1$, and that means that $2>x$. Now we have that $2>x>0$ which gives us that $x=1$. Putting that in $y+1=p^{x-1}$ we get that $y=0$, which is not an invalid solution, so p has to be 2 and from there we get that y=1 again.

Solutions: $(x,y,p)={(0,0,p),(1,1,2),(2,2,3)}$

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